javascript - 如何使用 RxJS 以适当的格式返回 observable 以及额外的单独分页数据

Question

我有一个返回以下内容的 API:

{
  'data' : {
     players: [
        {id: 1, name: 'harry'},
        {id: 2, name: 'barry'} ...
     ],
  },
  'paging': {
    current: 1,
    totalPages: 10,
  } 
}

如何正确返回对象中的玩家以及分页信息，以便它们是单独的键？

getPlayers(type: string): Observable<Player[]> {
  return this.get(myApiURL).pipe(
    map(result => {
      'players': result.data.map(player => {
        player._dead = !!player.deathDate;
        return player;
      }),
    }
  );
);

Answer 1

接口(interface)

interface Player {
  id;
  name;
}

interface Paging {
  current;
  totalPages
}

interface HttpPlayer  {
  data: {
    players: Player[]
  };
  paging: Paging;
}

interface UiPlayer {
  id;
  name;
  current;
  totalPages;
}

映射以合并播放器和寻呼(关于您的问题)

const mergePlayersWithPaging = (data: HttpPlayer): UiPlayer [] => data.players.map(player => Object.assign({}, player, data.paging))

映射到单独的播放器和寻呼(关于您的评论)

const mapHttpPlayerToPlayers = (httpPlayer: HttpPlayer): Player[] => httpPlayer.data.players;

const mapHttpPlayerToPaging = (httpPlayer: HttpPlayer): 

使用映射(合并)

getPlayers(type: string): Observable<Player[]> {
 return this.get(myApiURL).pipe(
   map(mergePlayersWithPaging)
 )
} 

使用映射(单独)

getPlayers(type: string): Observable<Player[]> {
 return this.get(myApiURL).pipe(
   map(mapHttpPlayerToPlayers)
 )
}

getPaging(type: string): Observable<Player[]> {
 return this.get(myApiURL).pipe(
   map(mapHttpPlayerToPaging)
 )
}